General College Chemistry Problem Set 20

Kinetic Molecular Theory of Gases 1. (a) What is meant by the root-mean-square (RMS) speed of gas particles? The root-mean-square (RMS) is the speed of a particle in a gas that has the average kinetic energy of all the particles of the sample (b) Suppose you are at a track where 7 joggers are doing laps. Their speeds are (in mph) 5, 5, 4, 3, 7, 8, and 9 mph. Calculate their average speed, then calculate their RMS speed (use 3 sig figs). Average speed: 5+5+4+3+7+8+9/ 7 = 5.86 mph RMS is the square root of the average of the squared speeds of all the particles in a gas Square root of 5^2 + 5^2 +4^2 + 3^2 + 7^2 + 8^2 + 9^2 / 7 = 38.4 mph 2. (a) Does pressure affect the root-mean-square speed of the particles in a gas? Why or why not? 3RT/ M, Pressure does not appear in the equation, therefore pressure has no effect (b) How does the RMS speed of a N2 molecule in a gas sample change under the following conditions (assuming the gas behaves ideally): a) Increasing the temperature Increasing the temperature increases the rms velocity of molecules, but it is not a direct proportionality. Rms velocity squared is proportional to the temperature b) Increasing the volume Increasing the volume while holding the temperature constants does not affect the rms velocity c) Adding Ar gas to the container while maintaining the same T. Mixing gases together at the same temp does not change rms velocity because the rms velocity only depends on T and the molar mass of the gas in question. Intermolecular forces are negligible for ideal gases (c) A sample of O2 gas is compressed by transferring it to a smaller container while keeping the T constant. What change do you expect in d) the average kinetic energy of the molecules, the average kinetic energy only depends on the T of the gas, so maintaining a constant T ensure that the average kinetic energy does not change e) the average speed of the molecules, the RMS speed of the gas only depends on the molar mass of the molecule and the temperature of the gas. Because T is constant, compressing the gas has no effect on the average speed f) the number of collisions that the molecules make with the container walls per unit time. Because the molecules are moving with the same average speed, but they are not in a smaller volume, they are expected to collide with the walls of the container more frequently g) the pressure of the container p is inversely proportional to V. Therefore, P should increase 3. (a) Rank the gases NO, NO2, N2O4, and N2O5 in order of increasing root-mean-square speed at 0°C. The lower the molar mass the greater the root mean square speed. The molar masses of NO, NO2, N2O4 are, and N2o5 are 30, 46, 92, 108 g/mol. Rank = N205, N204, NO2, NO (b) Below is the distribution of molecular speeds of CO2 and SO2 molecules at 25°C. Which curve is the profile for SO2? Which of the profiles should match that of propane (C3H8), a common fuel in portable grills? Explain your reasoning. 1 2 The large the molar mass, the lower the rms speed. Curve 1 shows more molecules at lower speed and curve 2 represents fewer molecules at higher speed. The molar masses of CO2 SO2 and C3H8 are 44, 64, and 44 g/mol. Curve 1 represents so2 with higher molar mass curve 2 represents both c02 c3h8 with the same molar masses 4. (a) Gases A and B have molecular weights of 65.2 g/mol and 101.2 g/mol, respectively. What is the rms velocity of these molecules at 23 ºC? (just for fun, you can convert their speeds to mph to see if these molecules would get a ticket on the freeway) 3(8.314 J/ mol K) (296 K) / 0.0652 kg/ mol square root = 337 m/s 3(8.314 J/ mol K) (296 K) . 0.1012 kg/ mo square root = 270 m/s (b) Suppose you have a container of gas with a molecular weight of 16.48 g/mol. If the RMS speed is 852.4 m/s, what is the temperature of the gas? T = urms 2 M/3R = 852.4^2 x 0.01649 . 3x8.314 = 480 K (c) 10.122 Enriching Uranium The two isotopes of uranium, 238U and 235U, can be separated by diffusion of the corresponding UF6 gases. What is the ratio of the root-mean-square speed of 238UF6 to that of 235UF6 at constant temperature? Square root of 349/352 = 0.996 5. (a) How is the rate of effusion of a gas related to its: (a) molar mass; (b) root-mean-square speed; (c) temperature? a. Rate of effusion is directly proportional to root mean squared. As the molar mass increases u RMS decreases because of the inverse relationship and the rate of effusion will decrease b. With the increase in urms the rate of effusion will also increase because of the direct relationship c. As the temp increases urms increase resulting in increase of rate of effusion (b) Which of the two outcomes shown below more accurately illustrates the effusion of gases from a balloon at constant atmospheric pressure if the red spheres have a greater root-mean-square speed than the blue spheres? Explain your reasoning. The red spheres effuse at a faster rate than blue spheres because of higher root mean square speed, therefore lost of red spheres should occur at a greater speed. Therefore, balloon a) is the right illustration as it loses 4 red spheres as opposed to b) which loses just 1 (a) (b) 6. (a) At a given temperature, gas A diffuses twice as fast as gas B. What is the relationship between the molar masses of the two gases? 2 squaring both sides = 4 (b) An unknown gas diffuses 3.728 times slower than hydrogen gas (H2) at any given temperature. What is the molecular weight of the unknown gas? Use Graham’s law of effusion/diffusion Mx/ mhs = 3.728 Mx = 3.728^2 x mh2 = (3.728)^2 (2.2016 g/mol) = 28.02 g / mol 7. Compounds sensitive to oxygen are often manipulated in glove boxes that may contain an atmosphere of pure nitrogen or pure argon. A rubber balloon filled with carbon monoxide was placed in such a glove box. After 24 hours, the volume of the balloon was unchanged. Did the glove box contain N2 or Ar? N2 The molar masses of the gasses are co-28 n2 -28 ar -40, because co and n2 have the same molar masses, they would effuse at the same rate and the volume won’t change 8. (a) Describe the difference between force and pressure. Force is the product of the mass of an object and acceleration due to gravity. Pressure is the force exerted by ban object over a given area F= mass x acceleration P= force/ area (b) Why is it easier to travel over deep snow when wearing the snowshoes shown to the right rather than just boots? The snow boots would distribute force due to mass of the person over a large surface area, reducing the pressure. This allows walking on the top of the snow without sinking (c) Why does an ice skater exert more pressure on ice when wearing newly sharpened skates than when wearing skates with dull blades? The sharp blades would exert force over small area increasing the pressure 9. Convert the following pressures to the indicated units: a) 694 torr in Pa 694 torr . 133.32 pa/ 1 torr = 92524 Pa b) 3.5 atm in torr 3.5 at, 760 torr/ 1 atm = 2660 torr c) 5.0  106 Pa in atm 5.0 . 10^6 pa . 9.87 . 10^-6 atm / 1 pa = 49.35 atm d) An official NCAA basketball is inflated to a pressure of about 11 psi above the normal 15 psi of atmospheric pressure. Express the total pressure in (a) atm, (b) torr, (c) Pascals The total pressure is 11 + 15 psi = 26 psi (1 atm / 14.7 psi) = 1.8 atm (760 torr/14.7 psi) + 1.3 x 10^3 torr e) Record High Atmospheric Pressure The highest atmospheric pressure ever recorded on Earth was 108.6 kPa at Tosontsengel, Mongolia, on December 19, 2001. Express this pressure in (a) millimeters of mercury, (b) atmospheres, and (c) millibars. a. 108.6 kPa x 1 atm/ 101.325 kPa x 760 nm hg/ 1 atom = 814.6 mmHg b. 108.6 kPa x 1 atom/ 101.325 kPa = 1.072 atm c. 108.6 kPa x 10 mbar/ 1 kPa = 1086 mbar f) Record Low Atmospheric Pressure Despite the destruction from Hurricane Katrina in August, 2005, the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered an atmospheric pressure of 88.2 kPa on October 19, 2005, about 2 kPa lower than Hurricane Katrina. What was the difference in pressure between the two hurricanes in (a) millimeters of mercury, (b) atmospheres, and (c) millibars? a. 2 kPa x 1 atm / 101.325 kPa x 760 mm Hg / 1 atm = 15 mm Hg b. 2kPa x 1 atm/ 101.325 kPa = 0.020 atm c. 2kPa x 10 mbar/ 1kPa = 20 mbar